hobomoe
Jun6-11, 07:16 PM
1. The problem statement, all variables and given/known data
Acidified potassium dichromate is an oxidant. It will react with titanium metal which is oxidised to form titanium ions.
Show that Ti^3+ and not Ti^2+ is formed in the reaction. Your answer should include chemical equations for any possible reactions and justification for you conclusion based of E values.
E(Cr207^2-/Cr^3+)=+1.33v E(Ti^2+/Ti)=-1.63v E(Ti^3+/Ti^2+)=-2.32v
2. Relevant equations
E(Cr207^2-/Cr^3+)=+1.33v E(Ti^2+/Ti)=-1.63v E(Ti^3+/Ti^2+)=-2.32v
3. The attempt at a solution
E(Cr207^2-/Cr^3+)-E(Ti^2+/Ti)=2.96v
E(Cr207^2-/Cr^3+)-E(Ti^3+/Ti^2+)=3.65v
Dichromate ions react with the titanium metal to form Ti^2+ ions. These ions then react with remaining dichromate ions to form Ti^3+ ions. It will try to form Ti^3+ ions from the Ti^2+ ions in order to get a greater E value.
Acidified potassium dichromate is an oxidant. It will react with titanium metal which is oxidised to form titanium ions.
Show that Ti^3+ and not Ti^2+ is formed in the reaction. Your answer should include chemical equations for any possible reactions and justification for you conclusion based of E values.
E(Cr207^2-/Cr^3+)=+1.33v E(Ti^2+/Ti)=-1.63v E(Ti^3+/Ti^2+)=-2.32v
2. Relevant equations
E(Cr207^2-/Cr^3+)=+1.33v E(Ti^2+/Ti)=-1.63v E(Ti^3+/Ti^2+)=-2.32v
3. The attempt at a solution
E(Cr207^2-/Cr^3+)-E(Ti^2+/Ti)=2.96v
E(Cr207^2-/Cr^3+)-E(Ti^3+/Ti^2+)=3.65v
Dichromate ions react with the titanium metal to form Ti^2+ ions. These ions then react with remaining dichromate ions to form Ti^3+ ions. It will try to form Ti^3+ ions from the Ti^2+ ions in order to get a greater E value.